Solving $\lim_{x\to0}\frac{ 5^x-3^x}{x}$ without L'Hôpital's rule The question is 
$$\lim_{x\to0}\frac{ 5^x-3^x}{x}$$
Unfortunately, we cannot use L'Hôpital's rule to solve this limit, so I'm not sure how to proceed with the question.
I'm sure there is an elegant way to manipulate the algebra to resolve the indeterminate form, however I just can't find it. Any help would be much appreciated thanks again.
 A: Remember the fundamental limits $\lim_{x \rightarrow 0} \frac{a^x -1}{x} = \ln(a)$. 
Then write $\frac{5^x - 3^x}{x} = \frac{5^x - 1}{x} - \frac{3^x -1}{x}$. 
So you have that $\lim_{x \rightarrow 0}\frac{5^x - 1}{x} - \frac{3^x -1}{x} = \lim_{x \rightarrow 0}\frac{5^x - 1}{x} - \lim_{x \rightarrow 0}\frac{3^x - 1}{x} = \ln(5) - \ln(3) = \ln(\frac{5}{3}) $
A: Let us consider the most general case of $$\lim_{x\to0}\frac{ a^x-b^x}{x}$$ $$\frac{ a^x-b^x}{x}=\frac{ e^{x \log(a)}- e^{x \log(b)}}{x} $$ For any value of $t$ we have $$e^t=\sum_{i=0}^\infty \frac{t^n}{n!}$$ So, replace $t$ by $x\log(a)$ and later by $x\log(b)$. This makes $$a^x-b^x=\left(1+x \log (a)+\frac{1}{2} x^2 \log ^2(a)+O\left(x^3\right) \right)-\left(1+x \log (b)+\frac{1}{2} x^2 \log ^2(b)+O\left(x^3\right) \right)$$ $$a^x-b^x=x (\log (a)-\log (b))+\frac{1}{2} x^2 \left(\log ^2(a)-\log
   ^2(b)\right)+O\left(x^3\right)$$ $$\frac{ a^x-b^x}{x}=(\log (a)-\log (b))+\frac{1}{2} x \left(\log ^2(a)-\log
   ^2(b)\right)+O\left(x^2\right)$$ 
$$\frac{ a^x-b^x}{x}=\log \left(\frac{a}{b}\right)\left(1+\log(\sqrt{ab})x\right)+O\left(x^2\right)$$ 
which shows the limit and also how it is approached when $x\to 0$.
